Make it clear to use `object_list` to get the QuerySet results in the template/view approach

[imran-uk]

Make it clear to use `object_list` to get the QuerySet results in the template/view approach

[JohnPaulGummapu]

@imran-uk
You're correct that it's not enough to pass page_obj alone to the template

When using Paginator in Django templates, you need to follow these steps.

-> In your view, create an instance of the Paginator class and use it to paginate your queryset.

from django.core.paginator import Paginator

def my_view(request):
    items_per_page = 10  # Define the number of items per page
    queryset = YourModel.objects.all()
    paginator = Paginator(queryset, items_per_page)
    
    page_number = request.GET.get('page')
    page_obj = paginator.get_page(page_number)
    
    results_list = page_obj.object_list  # Extract the list of objects for the current page
    
    context = {
        "page_obj": page_obj,
        "results_list": results_list,
    }
    
    return render(request, 'your_template.html', context)

-> In your template, you can then iterate over the results_list to display the items on the current page.

<ul>
    {% for item in results_list %}
        <li>{{ item.name }}</li>
        <!-- Display other item attributes as needed -->
    {% endfor %}
</ul>

->To create pagination controls, you can use the page_obj provided by the Paginator:

<div class="pagination">
    <span class="step-links">
        {% if page_obj.has_previous %}
            <a href="?page=1">&laquo; first</a>
            <a href="?page={{ page_obj.previous_page_number }}">previous</a>
        {% endif %}

        <span class="current-page">{{ page_obj.number }}</span>

        {% if page_obj.has_next %}
            <a href="?page={{ page_obj.next_page_number }}">next</a>
            <a href="?page={{ page_obj.paginator.num_pages }}">last &raquo;</a>
        {% endif %}
    </span>
</div>

[sathyaar5]

Hey @imran-uk that's correct!
It's important to note that you can't simply pass page_obj to the context and expect it to work like in the ListView approach.

Instead, explicitly extract the object_list like this:

page_obj = paginator.get_page(page_number)
results_list = page_obj.object_list

Then, include both "page_obj" and "results_list" in your context:

{
 "page_obj": page_obj,
 "results_list": results_list,
}

In your template, iterate over results_list to display the paginated results. This clarification can save you from the frustration of not seeing results when mistakenly passing just page_obj to your template.

[imran-uk]

Thank you for the explanation and detailed reply. I did figure it out eventually. What I'm saying is that the examples in the documentation need to be clear - your example below would be a good starting point, just add it to the page.

…

On Thu, Oct 12, 2023, 1:46 PM Sathya Narayana Rangappagari < ***@***.***> wrote: Hey @imran-uk <https://github.com/imran-uk> that's correct! It's important to note that you can't simply pass *page_obj* to the context and expect it to work like in the *ListView* approach. Instead, explicitly extract the *object_list* like this: page_obj = paginator.get_page(page_number) results_list = page_obj.object_list Then, include both "page_obj" and "results_list" in your context: `{ "page_obj": page_obj, "results_list": results_list, } ` In your template, iterate over *results_list* to display the paginated results. This clarification can save you from the frustration of not seeing results when mistakenly passing just *page_obj* to your template. — Reply to this email directly, view it on GitHub <#1413 (comment)>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AAEEOQXXVHC3H2K3MTFBZNTX67RCTAVCNFSM6AAAAAA5OHR6SWVHI2DSMVQWIX3LMV43OSLTON2WKQ3PNVWWK3TUHMYTONJZGU2DOMJTGE> . You are receiving this because you were mentioned.Message ID: ***@***.***>

[jackoconnordev]

If this issua still exists should a new issue be opened in the main Django repository, since the documentation pages are generated by Sphinx from docs in django/django?